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A coin is thrown until two heads and two tails appears. Let $Y$ be the number of throws until this happens. What is the cumulative distribution function of $Y$?

What I have gotten so far:

The last throw can end up being either head or tail. Let's look at this situations separately.

Last throw is tail: $$\sum^n_{k=2}P(Y=k)k=0,5^k\cdot k(k-1), \qquad n\ge4.$$

This happens to also be the probability mass function when the last throw is head. Now summing up the two cases, we get

$$0,5^k\cdot k(k-1)+0,5^k\cdot k(k-1)=2k(k-1)\cdot0,5^k$$

which is the probability mass function of the situation. Does this seem correct?

oooo
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    What does $k(k-1)$ stay for? – user Jul 13 '20 at 15:59
  • I think it stands for the toss happening twice witch is the requirement @user – BriggyT Jul 13 '20 at 18:22
  • I assume $0, 5$ means $1/2$ (i.e., $0.5$). But in that case your equation $\sum_{k=2}^n P[Y=k]k = (1/2)^k k(k-1)$ does not make sense because the left-hand-side is a function of $n$ while the right-hand-side is a function of $k$. Also I believe you want to compute $P[\mbox{exactly one tail in $n-1$ tosses, then $n$th toss tail}]$ for your left-hand-side (assuming $n-1\geq 3$). I do not know where the expression $\sum_{k=2}^n P[Y=k]k$ comes from, if it is somehow the same then that is a big unexplained jump. – Michael Jul 13 '20 at 18:36

1 Answers1

2

Hint:

The probability that the $k$-th throw is the last one is the probability of the event that

  1. $k>3$

and

either

  1. there is exactly one T in the previous $k-1$ throws and the $k$-th throw is T,

or

  1. there is exactly one H in the previous $k-1$ throws and the $k$-th throw is H.

This is: $$ \binom{k-1}1\frac1{2^{k-1}}\cdot\frac12+\binom{k-1}1\frac1{2^{k-1}}\cdot\frac12=\binom{k-1}1\frac1{2^{k-1}}, $$ where $\binom{k-1}1$ is the number of ways to order 1 T(H) and $k-2$ H(T).

Particularly this consideration proves: $$ \sum_{k=3}^\infty\frac k{2^k}=1. $$

user
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